If A is a square matrix of order 3 such that `A("adj A")=[(-3,0,0),(0,-3,0),(0,0,-3)]`, then `|A|` is equal to

Options

A. 9
B. `-3`
C. `-6`
D. 3

Answer

To solve the problem, we need to find the determinant of the square matrix \( A \) of order 3 given that \( A \cdot \text{adj}(A) = \begin{pmatrix} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Understanding the given equation**:
We start with the equation:
\[ A \cdot \text{adj}(A) = \begin{pmatrix} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{pmatrix} \]
This can be rewritten as:
\[ A \cdot \text{adj}(A) = -3I \]
where \( I \) is the identity matrix of order 3.

2. **Using the property of determinants**:
We know from linear algebra that:
\[ A \cdot \text{adj}(A) = \det(A) \cdot I \]
Therefore, we can equate:
\[ \det(A) \cdot I = -3I \]

3. **Equating the determinants**:
From the above equation, we can deduce that:
\[ \det(A) = -3 \]

4. **Finding the determinant of the adjoint**:
We also know that for any square matrix \( A \) of order \( n \):
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \]
Here, \( n = 3 \), so:
\[ \det(\text{adj}(A)) = \det(A)^{3-1} = \det(A)^2 \]

5. **Calculating the determinant of the adjoint**:
Substituting \( \det(A) = -3 \):
\[ \det(\text{adj}(A)) = (-3)^2 = 9 \]

6. **Final conclusion**:
Therefore, the value of \( |A| \) or \( \det(A) \) is:
\[ \det(A) = -3 \]

### Summary:
The determinant \( |A| \) is equal to \( -3 \).